The equation for the reaction is

H_{2}SO_{4} + 2 LiOH ---> Li_{2}SO_{4} + 2 H_{2}O

We have n(H_{2}SO_{4}) = cV = 0.0500 x 50.00 x 10^{-3} = 2.50 x 10^{-3} mol

And n(LiOH) = cV = 0.200 x 30.00 10^{-3} = 6.00 x 10^{-3} mol

And n(LiOH) used to neutralise H_{2}SO_{4} = (2 / 1) x 2.50 x 10^{-3} = 5.00 x 10^{-3} mol

So, n(LiOH) after reaction = n(LiOH) initially - n(LiOH) reacted = 6.00 x 10^{-3} - 5.00 x 10^{-3} = 1.00 x 10^{-3} mol

So, after reaction, [OH^{-}] = [LiOH] = n / V = 1.00 x 10^{-3} / [(50.00 + 30.00) x 10^{-3}] = 0.0125 mol L^{-1} (3 sig fig)

So, assuming the temperature is 25 degC, pH = 14 + log_10 [OH-] = 12.097 (3 dec. pl.)

So, the answer is **B**.

The above response has the reaction of sulfuric acid and lithium hydroxide as 1:1, leading to an incorrect answer.